arX

iv:1

409.

3114

v2 [

mat

h.N

T]

17

Feb

2016

C’est pour toi que je joue, Alf c’est pour toi,Tous les autres m’ecoutent, mais toi tu m’entends ...Exile d’Amsterdam vivant en Australie,Ulysse qui jamais ne revient sur ses pas ...Je suis de ton pays, meteque comme toi,Quand il faudra mourir, on se retrouvera1

To the memory of Alf van der Poorten

ON THE VANISHING OF IWASAWA’S CONSTANT µ FOR

THE CYCLOTOMIC Zp-EXTENSIONS OF CM NUMBER

FIELDS.

PREDA MIHAILESCU

Abstract. We prove that µ = 0 for the cyclotomic Zp-extensions ofCM number fields.

Contents

1. Introduction 11.1. Notations and basic facts on decomposition of Λ-modules 21.2. Decomposition and Thaine Shifts 31.3. Constructing the base field 51.4. Plan of the paper 72. Thaine shift and proof of the Main Theorem 82.1. The vanishing of µ 103. Appendix 103.1. Auxiliary facts 103.2. Applications of the Tchebotarew Theorem 113.3. Proof of Proposition 2.3 and related results 123.4. The Norm Principle, ray class fields and proof of Lemma 2.2 153.5. Proof of the Proposition 1.3 19References 23

1. Introduction

Iwasawa gave in his seminal paper [4] from 1973 examples of Zp-extensionsin which the structural constant µ 6= 0. In the same paper, he proved that

1Free after Georges Moustaki, “Grand-pere”Date: Version 2.0 August 18, 2018.Key words and phrases. 11R23 Iwasawa Theory, 11R27 Units.

1

2 PREDA MIHAILESCU

if µ = 0 for the cyclotomic Zp-extension of some number field K, then theconstant vanishes for any cyclic p-extension of K – and thus for any numberfield in the pro-p solvable extension of K. Iwasawa also suggested in thatpaper that µ should vanish for the cyclotomic Zp-extension of all numberfields, a fact which is sometimes called Iwasawa’s conjecture. The conjecturehas been proved by Ferrero and Washington [3] for the case of abelian fields.In this paper, we give an independent proof, which holds for all CM fields:

Theorem 1.1. Let K be a CM number field and p an odd prime. Then

Iwasawa’s constant µ vanishes for the cyclotomic Zp-extension K∞/K.

1.1. Notations and basic facts on decomposition of Λ-modules. Inthis paper p is an odd prime. In the sequel, the Iwasawa constant µ for somenumber field K will always refer to the µ-invariant for the Zp-cyclotomic ex-tension of K. We shall denote number fields by black board bold characters,e.g. K,L, etc. If K is a number field, its cyclotomic Zp-extension is K∞ andB∞/Q is the Zp-extension of Q, so K∞ = K · B∞. We denote as usual by Γthe Galois group Gal(K∞/K) and let τ ∈ Γ be a topological generator.

Let B1 = Q and Bn ⊂ B∞ be the intermediate extensions of B∞ and letthe counting be given by [Bn : Q] = pn−1, so Bn = B∞∩Q[ζpn]. Let κ > 0 bethe integer for which K ∩ B∞ = Bκ; then µpκ ⊂ K and µpκ+1 6⊂ K. We shalluse a counting of the intermediate fields of K∞ that reflects this situationand let K0 = K1 = . . . = Kκ = K and [Kκ+1 : K] = p, etc. The constantκ can be determined in the same way for any number fields and we adoptthe same counting in any cyclotomic Zp-extension. This way, for n ≥ κ wealways have µpn ⊂ Kn. We let γ ∈ Gal(B∞/Q) be a topological generatorof Gal(B∞/Q) and let τ ∈ Γ := Gal(K∞/K) be a topological generator for

Γ. We may thus assume that τ = γpκ−1

for some lift γ ∈ Gal(K∞/Q) of γand write as usual T = τ − 1,Λ = Zp[[T ]] and

ωn = τpn−κ − 1 = (T + 1)p

n−κ − 1,

νm,n = ωm/ωn, for m > n ≥ κ.Since K = K1 = Kκ, we may also write νm,1 = νm,κ. Note that the specialnumeration of fields which we introduce in order to ascertain that µpn ⊂ Kn

induces a shift in the exponents in the definition of ωn. In terms of γ werecover the classical definitions, but our exposition is made in terms of thetopological generator τ . Note that the base field K with respect to whichwe shall bring our proof, is still to be defined, and will be a CM on in whichwe assume that µ > 0, and in which some useful additional conditions arefulfilled.

The p-Sylow subgroups of the class groups C(Kn) of the intermediatefields of Kn are denoted as usual by An = An(K) = (C(Kn))p; the explicitreference to the base field K will be used when we refer simultaneouslyto sequences of class groups related to different base fields. Traditionally,A = A(K) = lim−→An. We use the projective limit, which we denote by

VANISHING OF µ 3

pA−(K) = lim←−(An), as explained in detail below. The maximal p-abelian

unramified extensions of Kn are denoted by Hn = Hn(K) ⊃ Kn and Xn =Gal(Hn/Kn). The projective limit with respect to restriction maps is X =lim←−(Xn): it is a Noetherian Λ-torsion module on which complex conjugation

acts inducing the decomposition X = X+ ⊕ X−. The Artin maps ϕ :An → Xn are isomorphisms of finite abelian p-groups; whenever the abelianextension M/K is clear in the context, we write ϕ(a) for the Artin Symbol(M/Ka

), where a ∈ C(K) if M is unramified, or a is an ideal of K otherwise.

Complex conjugation acts on class groups and Galois groups, inducing directsum decompositions in plus and minus parts:

An = A+n ⊕A−

n , Xn = X+n ⊕X−

n , etc.

The idempotents generating plus and minus parts are 1±2 ; since 2 is a

unit in the ring Zp acting on these groups, we also have Y + = (1 + )Yand Y − = (1 − )Y , for Y ∈ {An,Xn,X, . . .}. Throughout the paper,we shall use, by a slight abuse of notation and unless explicitly specifiedotherwise, the additive writing for the group ring actions. This is prefer-able for typographical reasons. If M is some Noetherian Λ-torsion mod-ule on which acts, inducing a decomposition M = M+ ⊕M−, we writeµ−(M) = µ(M−), λ−(M) = λ(M−), etc, for the Iwasawa constants of thismodule. In the case when M = pA−(M) or M = X(M) is attached to somenumber field M, we simply write µ−(M) or µ(pA−(M)).

By assumption on K, the norms NKm/Kn: Am → An are surjective for all

m > n > 0. Therefore, the sequence (An)n∈N is projective with respect tothe norm maps and we denote their projective limit by pA = lim←−n

An. The

Artin map induces an isomorphism of compact Λ-modules ϕ : pA→ X. Theelements of pA are norm coherent sequences a = (an)n∈N ∈ pA with an ∈ An

for n ≥ κ; we let a0 = a1 = . . . = aκ. It is customary to identify X with pAvia the Artin map; we shall not use injective limits here, but make explicitreference to the projective limit pA.

It is a folklore fact that if µ vanishes for the cyclotomic Zp-extensionof some CM field Kstart, then it vanishes for any finite algebraic extensionthereof. We prove this in Fact 3.1 in the Appendix. In order to prove theTheorem 1.1 we shall need to taylor some base field K/Kstart, which is aGalois CM extension of Q and enjoys some additional conditions that shallbe discussed bellow; of course, we assume that µ(K) > 0. Before describingthe construction of K however, we need to introduce some definitions andauxiliary constructions.

1.2. Decomposition and Thaine Shifts. Let M be a Noetherian Λ-torsion module. It is associated to an elementary Noetherian Λ-torsionmodule E(M) ∼M defined by:

E(M) = Eλ(M) ⊕ Eµ(M), with

Eµ(M) = ⊕ri=1Λ/(p

eiΛ), Eλ(M) = ⊕r′j=1Λ/(f

e′jj Λ),

4 PREDA MIHAILESCU

where all ei, e′j > 0 and fj ∈ Zp[T ] are irreducible distinguished polynomials.

The pseudoisomorphism M ∼ E(M) is given by the exact sequence

1→ K1 →M → E(M)→ K2 → 1,(1)

in which the kernel and cokernel K1,K2 are finite. We define λ- and µ-partsof M as follows:

Definition 1.2. LetM be a Noetherian Λ-torsion module. The λ-part L(M)is the maximal Λ-submodule of M of finite p-rank. The µ-part M(M) is

the Zp-torsion submodule of M ; it follows from the Weierstraß Preparation

Theorem that there is some m > 0 such thatM(M) =M [pm]. The maximal

finite Λ-submodule of M is F(M), its finite part. By definition, L(M) ∩M(M) = F(M).

Let the module D(M) = L(M) +M(M) be the decomposed submoduleof M . Then for all x ∈ D(M) there are xλ ∈ L(M), xµ ∈ M(M) such

that x = xλ + xµ, the decomposition being unique iff F(M) = 0. The

pseudoisomorphism M ∼ E(M) implies that [M : D(M)] <∞.

If x ∈M \ D(M), the L- and the D-orders of x are, respectively

ℓ(x) = min{j > 0 : pjx ∈ L(M)}, and(2)

δ(x) = min{k > 0 : pkx ∈ D(M)} ≤ ℓ(x).We say that a Noetherian Λ-module M of µ-type is rigid if the map ψ :

M → E(M) is injective. Rigid modules have the fundamental property that

for any distinguished polynomial g(T ) ∈ Zp[T ] and any x ∈Mg(T )x = 0 ⇔ x = 0.(3)

Note that for CM base fields M, the modulesM(pA−(M∞)), where M∞

is the cyclotomic Zp-extension, are rigid. The following fact about decom-position is proved in the last section of the Appendix.

Proposition 1.3. Let M be a number field, let Tn ⊃ H(Mn) ⊃ Mn be the

ray class fields to some fixed ray M0 ⊂ O(M) and M = pA−(M) be the

projective limit of the galois groups Tn = Gal(Tn/Mn). Assume in addition

that the following condition is satisfied by M: if r = p-rk(L(M)) and L1 =NM∞/M1

(L(M)) then

p-rk(L1) = r and ord(x) > p2 for all x ∈ L1 \ Lp1.(4)

If these hypotheses hold and x ∈ M is such that px ∈ D(M), then T 2x ∈D(M).

We observe that the condition (4) can be easily satisfied by eventuallyreplacing an initial field M with some extension, as explained in Fact 3.2 ofAppendix 3.1

Next we define Thaine shifts and lifts. Let M be a CM Galois field andconsider a = (an)n∈N ∈ pA−(M), some norm coherent sequence. We assumethat the norm maps NMn/Mn′

: A−n (M) → A−

n′(M) are surjective for all

n > n′ ≥ 1 and fix some integer m > κ(M) – with M ∩ B∞ = Bκ(M) – and

VANISHING OF µ 5

a totally split prime q ∈ am, which is inert in M∞/Mm. Let q ∈ Q be therational prime below q and assume that q ≡ 1 mod p. Let F ⊂ Q[ζq] be thesubfield of degree p in the q-th cyclotomic extension. We let Ln = Mn · Fand L∞ = M∞ · F. The tower L∞/L is the inert Thaine shift of the initialcyclotomic extension M∞/M, induced by q ∈ am. Let Q ⊂ Lm be theramified prime above q. According to Lemma 3.5 in the Appendix, wemay apply Tchebotarew’s Theorem in order to construct a sequence b =(bn)n∈N ∈ pA−(L) such that bm = [Q] is the class of Q and NLn/Mn

(bn) = anfor all n ∈ N. In the projective limit, we then also have NL∞/M∞

(b) = a. Asequence determined in this way will be denoted a Thaine lift of a. It is notunique.

Let F = Gal(F/Q) be generated by ν ∈ F ; we write s := ν − 1 and

Φp(ν) =(s+1)p−1

s . By using the identity xp−1x−1 = (y+1)p−1

y = yp−1 +O(p), we

see that the algebraic norm verifies

N :=

p−1∑

i=0

νi = Φp(ν) = pu(s) + sp−1 = p+ sf(s),(5)

f ∈ Zp[X] \ pZp[s], ∈ (Z/(pN · Z)[s])×,∀N > 0.

Since q is totally split in Mm/Q, the extensions Ln/Q are Galois and Fcommutes with Gal(Mm/Q), for every m and Gal(Ln/Mn) ∼= F .

1.3. Constructing the base field. For our proof we shall choose a basefield K as follows. Start with some CM field Kstart for which one assumes

that µ > 0, and let −D be a quadratic non-residue modulo p – so(−Dp

)=

−1 – and k = Q[√−D] be a quadratic imaginary extension. Our start field

should be galois and contain the p-th roots of unity. We require thus that

K ⊃ K(n)start[ζp,

√−D].

We additionally expect that the primes that ramify in K∞/K be totallyramified and the norms Nn,m : A(Kn) → A(Km) be surjective, so K∞ ∩H(K) = K. We also require that the exponent exp(M(pA−(K))) ≥ p2, whichcan be achieved by means of Fact 3.2. The first step of the construction

consists thus in replacing Kstart by an initial field Kini = K(n)start[ζp,

√−D].

If exp(M(pA−(Kini))) = p, then replace Kini by some Thaine shift thereof, inorder to increase the exponent. We need to fulfill the condition (4) requiredin Proposition 1.3; for this we determine an integer t as follows: let L =L(pA−(Kini)) and r = p-rk(L) = λ(Kini). Let Lt = NKini,∞

/Kini;t(L). Wethen require that

p-rk(Lt) = r and ord(x) > p2 for all x ∈ Lt \ Lpt .(6)

With this, we let K′ini = Kini,t ⊂ Kini,∞. Finally we apply the Proposition

1.3 in order to choose a further extension of K′ini in the same cyclotomic Zp-

extension, that yields some simple decomposition properties for Thaine lifts.

6 PREDA MIHAILESCU

Q

k = Q(√−D)

Kstart

Kini = K(n)start(ζp,

√−D)

K′ini = Kini,t

Bκ

B

kκ

k∞

K = K′ini,k′ = Kini,k′+t

K∞

Figure 1. Construction of the base-field K

Let κ′ = κ(K′ini) and τ = γp

κ′−1

generate Γ := GalK′ini,∞/K

′ini. Recall

that γ is a generator of Gal(B∞/B), which explains the definition of τ ; finally,

T = τ−1 and we let pb be the exponent ofM(pA−(K′ini)). We let k′ be such

that ωk′ ∈ (pb+1, T 2(b+1)) and define K = K′ini,k′ = Kini,k+t and let finally

κ = κ(K) and τ = γpκ−1

, T = τ − 1, etc. We conclude from Proposition 1.3and the choice of k′ that

T · pA−(K) ⊂ D(pA−(K)).(7)

Moreover:

Remark 1.4. Suppose that L/K is a Thaine shift and y ∈ pA−(L) is such

that either

1. py = x+ w with x ∈ ιL/K(pA−(K)) and w ∈M(pA−(L)), or

2. pb+1y ∈ L(pA−(L)).

Then Ty ∈ D(pA−(L)) too.

The second point is a direct consequence of the choice of k′ and of Propo-

sition 1.3. For the first, since x ∈ pA−(K), we know that pbx ∈ L(pA−(K)),so pb+1y = pbx− pbw ∈ L(pA−(K))+M(pA−(L)) ⊂ D(pA−(L)), and the fact

follows from point 2.

The construction of the Thaine shift is shown in the Figure 2

VANISHING OF µ 7

Q F Q(ζq)

K L

B

Km

K∞

Lm

L∞

p

Figure 2. Thaine shift and lift

This concludes the sequence of steps for the construction of the base fieldK, which are reflected in the figure 1. We review the conditions fulfilled bythis field:

0. The field K′ = K(n)start[ζp,

√−D] and Kini = K′

s with t subject to (6).1. The field K is a Galois CM extension K/Q which contains the p-th

roots of unity and such that µ(K) > 0 for the cyclotomic Zp-extensionof K. The primes that ramify in K∞/K are totally ramified.

2. We have T · (pA−(K)) ⊂ D(pA−(K)) and the properties in Remark1.4 are verified.

3. The numbering of intermediate fields starts from κ, where K ∩ B =Bκ.

4. The exponent exp(M(pA−(K))) ≥ p2.5. The fieldK contains an imaginary quadratic extension k = Q[

√−d] ⊂

K which has trivial p-part of the class group.

1.4. Plan of the paper. We choose a base field K as shown in the previoussection and a norm coherent sequence

a = (an)n∈N ∈ M(pA−(K)) \(p · pA−(K) + (p, T )M(pA−(K))

).

We note that condition 0. in the choice of Kini readily implies that ord(a) =ord(a1), so (ord(a)/p) · a1 6= 0 and thus

(ord(a)/p) ∈ M(pA−(K))[p] \ TM(pA−(K))[p].(8)

We let o := oT (a) ≥ 0 be such that a ∈ (T o) · pA−(K) \ (T o+1) · pA−(K).In the second Chapter, we build a Thaine shift with respect to a prime

q ∈ am and a lift b to a and derive the main cohomological propertiesof the shifted extension. The most important facts are the decompositionTx ∈ D(pA−(L)) for all x ∈ pA−(L) with ℓ(x) ≤ p · ord(M(pA−(K)) and

vanishing of the Tate cohomology H0(F, pA−(L)). Based on this and the factthat Tb is decomposed while sbm = 0, as the class of a ramified ideal, weobtain in Chapter 3 a sequence of algebraic consequences which eventually

8 PREDA MIHAILESCU

lead to the fact that Ta = N (b) ∈ ωmM(pA−(K)); since this holds forarbitrary choices of m, independently of a, we obtain a contradiction to (8),which proves the Iwasawa conjecture.

The paper is written so that the main ideas of the proof can be presentedin the main part of the text, leading in an efficient way to the final proof.The technical details and results are deduced with a richness of detail, inthe appendices.

2. Thaine shift and proof of the Main Theorem

We have selected in the first Chapter a base field K which is CM and en-dowed with a list of properties. Consider the Fp[[T ]]-module P := pA−(K))/(p)and let π : pA−(K) → P be the natural projection. Then for any a ∈M(pA−(K)) \ p · pA−(K) there is some integer oT (a) ≥ 0 such that theimage π(a) ∈ pA−(K)/(p) verifies π(a) ∈ T oT (a)π(pA−(K)). We choosea ∈ M(pA−(K)) \ p · pA−(K) with the minimal value of oT (a); let m > κ(K)be such that

deg(ωm) > 2(oT (a) + 1)(9)

and q ∈ am be a totally split prime. Let q ⊂ N be the rational prime belowq; since K contains the pm-th roots of unity, it follows that q ≡ 1 mod pm.We let L = K ·F;L∞ = K∞ ·F be the Thaine shift induced by q, as describedin the section §1.2 and let b ∈ pA−(L) be a Thaine lift of a.

For C some Zp[s]-module, we use the Tate cohomologies associated to C,defined by

H0(F,C) = Ker (s : C → C)/(NC),(10)

H1(F,C) = Ker (N : C → C)/(sC).

The notation introduced here will be kept throughout the paper.Let B′

n ⊂ A−(Kn) be the submodule spanned by the classes of primesthat ramify in Ln/Kn. By choice of L, these are the primes above q andconsequently B′

n = ιm,n(B′m) for all n > m. Here ιm,n : A−(Lm)→ A−(Ln)

is the natural lift map. We let pv be the exponent of B′m, so pvB′

n = 0 forall n ≥ m.

Since B′n is constant up to isomorphism for all n > m, the vanishing of

H0(F, pA−(L)) is a straight forward consequence of :

Lemma 2.1.

Ker (s : pA−(L)→ pA−(L)) = ι(pA−(K))(11)

In particular, H0(F, pA−(L)) = 0.

Proof. Consider x = (xn)n∈N ∈ Ker (s : pA−(L) → pA−(L)) and let N >

m+b. Let X ∈ xN be a prime: then (Xs(1−)) = (ξ1−), for some ξ ∈ LN andN (ξ1−) ∈ µ(KN ). Since qm is inert in KN/Km, we have N (LN )∩µ(KN ) ⊂µ(KN )p. We may thus assume, after eventually modifying ξ by a root of

VANISHING OF µ 9

unity, that N (ξ1−) = 1. Hilbert’s Theorem 90 implies that there is someα ∈ LN such that

Xs(1−) = (ξ1−) = (α1−)s ⇒ (X/(α))(1−)s = (1).

The ideal Y := (X/(α))1− ∈ x2N verifies Ys = (1). If xN 6∈ ι(pA−(K)),then Y must be a product of ramified primes, so x2N ∈ B′

N+ιKN ,LN(A−

N (K)).Recall that B′

N = ιm,N (B′m) is spanned by the classes of the ramified

primes and pvB′N = 0. In particular x2N ∈ B′

N + ιKN ,LN(A−

N (K)) andB′

N = ιm,N (B′m) imply that

xN−v = NN,N−v(xN ) ∈ ιKN−v,LN−v(A

−(KN−v))+B′m

pv= ιKN−v

,LN−v(A−(KN−v)).

This happens for all N > m+ v, so x ∈ ιK,L(pA−(K)), as claimed. �

Note that at finite levels we have

H0(F,A−n (L))

∼= B′n/(B

′n ∩ ιL/K(A−

n (K))) 6= 0.

The Herbrand quotient of finite groups is trivial, so |H0(F,A−n (L))| =

|H1(F,A−n (L))|. In the projective limit however, H1(F, pA−(L)) 6= 0 so

equality is not maintained. In the Appendix §3.3, we prove though thefollowing result:

Lemma 2.2. Suppose that v ∈ M(pA−(L)) has non trivial image in the

Tate group H1(F,

(M(pA−(L))

)). Then either ord(v) > p exp(M(pA−(K))

or T 2v ∈ spA−(L).

We turn now our attention to the decomposition of the Thaine lift b; weprove in the Appendix 3.3 the following

Proposition 2.3. Let F =< ν > be a cyclic group of order p acting on

the p-abelian group B and let x ∈ B be such that y = N (x) has order

ord(y) = q = pl > p. Then ord(x) ≤ pq.It implies

Corollary 2.4. Let y ∈ pA−(L) be such that ℓ(N (y)) = q > p. Then

q ≤ ℓ(y) ≤ pq and Ty ∈ D(pA−(L)). All these facts hold in particular for

any Thaine lift b. In this case, one has additionally ord(sb) ≤ ord(b) ≤ pq.Proof. Let f(T ) ∈ Zp[T ] be a distinguished polynomial that annihilates

pℓ(y)y ∈ L(pA−(L)) and let β = f(T )y. Then α := N (y) has the orderq > p, by hypothesis, so we may apply the Proposition 2.3. This impliesthat ord(β) ≤ pq and thus ℓ(y) ≤ pq ≤ p exp(M(pA−(K))), by definitionof this order. The first claim in Remark 1.4 implies that Ty ∈ D(pA−(L)).Since ord(a) = ord(N (b)) > p by choice of a, the statement applies inparticular to any Thaine lift b. In this case, we know that pbm = am andord(am) = q, hence ord(b) ≥ ord(bm) = pq, hence ord(b) = pq. We obviouslyhave ord(sb) ≤ ord(b). �

10 PREDA MIHAILESCU

2.1. The vanishing of µ. Since sbm = 0, the Theorem VI of Iwasawa[5] implies that there is some c ∈ pA−(L) such that sb = νm,1c. Thenνm,1(N (c)) = 0 and the Fact 3.4 in the Appendix implies that N (c) = 0.Moreover,

pℓ(b)c = pℓ(b)(νm,1sb) ∈ L(pA−(L))

so, by Corollary 2.4, ℓ(c) ≤ pq too, and thus Tc ∈ D(pA−(L)). Let Tc = cλ+cµ. Since L(pA−(K)) ∩M(pA−(K)) = 0, it follows that N (cλ) = N (cµ) = 0,individually. We have, by comparing parts, sbµ = νm,1cµ, so pq · νm,1cµ = 0,and sinceM(pA−(L)) is rigid, (3) implies that ord(cµ) ≤ pq. We may thusapply Lemma 2.2, which implies that T 2cµ ∈ sM(pA−(L)), say T 2cµ = sx;then s(T 2bµ−νm,1x) = 0 and Lemma 2.1 implies that T 2bµ = νm,1x+z, z ∈ι(pA−(L)). By taking norms we obtain T 3a = νm,1x + pz. This impliesoT (a) + 3 ≥ deg(νm,1) and we can choose m large enough, to obtain acontradiction. This confirms the claim of the Main Theorem.

3. Appendix

In the Appendix, unless otherwise specified, the notation used in thevarious Facts and Lemmata is the one used in the section where these areinvoked in the text. In the next section we provide a list of disparate, usefulfacts:

3.1. Auxiliary facts. We start by proving that if µ > 0 for some numberfield, then it is also non - trivial for finite extensions thereof.

Fact 3.1. Let K be a number field for which µ(K) 6= 0 and L/K be a finite

extension, which is Galois over K. Then µ(L) 6= 0.

Proof. If M ⊂ L has degree coprime to p, then Ker (ι : A(K)→ A(M)) =0, so we may reduce the proof to the case of a cyclic Kummer extension ofdegree p. Let M = LGal(L/K)p be the fixed field of some p-Sylow subgroupof Gal(L/K). Then p does not divide [M : K], so Ker (ι : A−(K) →A−(M)) = 0, and thus µ(M) 6= 0. We may assume without loss of generality,that M contains the p-th roots of unity. Since p-Sylow groups are solvable,the extension L/M arises as a sequence of cyclic Kummer extensions ofdegree p. It will thus suffice to consider the case in which k is a numberfield with µ 6= 0 and containing the p-th roots of unity and k′ = k[a1/p] is acyclic Kummer extension of degree p. We claim that under these premises,µ(k′) 6= 0. Let kn ⊂ k∞ and k′n ⊂ k′∞ be the intermediate fields of thecyclotomic Zp-extensions, let ν generate Gal(k′/k). Let F/k∞ be an abelianunramified extension with Gal(F/k∞) ∼= Fp[[T ]]; such an extension mustexist, as a consequence of µ > 0. There is thus for each n > 0 a δn ∈ k×nsuch that Fn = kn

[δFp[[T ]]/pn

]is an unramified extension with galois group

Gn = Gal(Fn/kn) of p-rank rn := p-rk(Gn) > pn−c for some c ≥ 0. We

define F ′n = Fn[a

1/p] and let F′n ⊃ F ′

n be the maximal subextension which

is unramified over k′n. We have F′n ⊇ Fn and thus p-rk(Gal(F

′n/k

′n)) ≥

VANISHING OF µ 11

p-rk(Gal(Fn/kn)) → ∞. Consequently, k′∞ has an unramified elementaryp-abelian extension of infinite rank, and thus µ(k′) > 0, which completesthe proof. �

Fact 3.2. Let K be a CM extension with µ > 0. Then it is possible to build

a further CM extension L/K with exp(M−(L)) > p2.

Proof. We have shown in Fact 3.1 that we may assume that µp3 ⊂ K. Let a =

(an)n∈N ∈ M−(K) and q ∈ a2, with a2 6= 0, be a totally split prime whichis inert in K∞/K2. Let L/K2 be the inert Thaine shift of degree p2 induced

by q, let b2 = [Q(1−)/2] be the class of the ramified prime of L above q andb = (bm)m∈N be a sequence through that extends b2, such that NL/K(b) = a.

Then b 6∈ L−(L) and there is some polynomial f(T ) ∈ Zp[T ] such thatf(T )b ∈ M−(L), while NL/K(f(T )b) = f(T )a. The capitulation kernel

Ker (ι : A−(Kn) → A−(L)) is trivial and consequently ord(Tf(T )b) ≥p2ord(a); hnece exp(M−(L)) > p2. Thus L verifies the claimed properties.

�

The following fact was proved by Sands in [6]:

Fact 3.3. Let L/K be a Zp-extension of number fields in which all the primes

above p are completely ramified. If F (T ) ∈ Zp[T ] is the minimal annihilator

polynomial of L(L), then (F, νn,1) = 1 for all n > 1.

As a consequence,

Corollary 3.4. Let K be a CM field and suppose that x ∈ pA−(K) verifies

νn,1x = 0. Then x = 0.

Proof. Let q be the exponent of the Zp-torsion of pA−(K). It follows thenthat qx ∈ L(pA−(K)) is annihilated by νn,1, so the Fact 3.3 implies thatqx = 0 and thus x ∈ M(pA−(K)). Since νm,1x = 0 it follows that x = 0, asclaimed. �

3.2. Applications of the Tchebotarew Theorem. We prove the exis-tence of Thaine lifts.

Lemma 3.5. Let K be a CM field and a = (an)n∈N ∈ pA−(K) and L = K ·Fbe a Thaine shift induced by a split prime q ∈ am. Then there is a Thaine

lift b = (bn)n∈N ∈ pA−(L) with the properties that N (b) = a and bm is the

class of the ramified prime Q ⊂ L above q.

Proof. We prove by induction that for each n ≥ 1 there is a class bn ∈ A−n (L)

with N (bn) = an and Nn,n−1(bn) = bn−1. The claim holds for n ≤ m bydefinition. Assume that it holds for n ≥ m and consider minus parts of themaximal p-abelian unramified extensions

H(Kn)−/Kn,H

−(Ln)/Ln,H−(Ln+1)/Ln+1, and H−(Kn+1)/Kn+1.

For H′ ∈ {H−(Kn),H−(Ln),H

−(Kn+1)}, we obviously have H′ · Ln+1 ⊂H−(Ln+1). For some unramified Kummer extension M/K we denote by

12 PREDA MIHAILESCU

ϕ(x) =(M/Kx

)the Artin symbol of a class or of an ideal. The induc-

tion hypothesis implies that both ϕ(bn) and ϕ(an+1) restrict to ϕ(an) ∈Gal(H−(Kn)/Kn), and since A−

n+1(K) and A−n (L) surject by the respective

norms to A−n (Kn), it follows that

Ln+1H(Kn+1) ∩ Ln+1H−(Ln) = Ln+1H

−(Kn).

There is thus some automorphism x ∈ Gal(H−(Ln+1)/Ln+1), such that

x∣∣Gal(H−(Ln)/Ln)

= ϕ(bn) and x∣∣Gal(H−(Kn+1)/Kn+1)

= ϕ(an+1).

By Tchebotarew, there are infinitely many totally split primes R ⊂ Ln+1

with Artin symbol(H−(Ln+1)/Ln+1

R

)= x, and by letting bn+1 = [R] for such

a prime, we have N (bn+1) = an+1 and Nn+1,n(bn+1) = bn. We obtain byinduction a norm coherent sequence b = (bn)n∈N which verifies bm = [Q]and N (b) = a, and this completes the proof. �

3.3. Proof of Proposition 2.3 and related results. Let the notationsbe like in the statement of the Proposition and let q = pk = ord(y) andr ≥ q be the order of x and R = Z/(r · Z)[s]. Then R has maximal ideal(p, s) and s is nilpotent. By definition, R acts on x and we consider themodules X = Rx and Y = Z/(q′ · Z)y ⊂ X.

With these notations, we are going to prove that pqx = 0. We start byproving

Lemma 3.6. Under the given assumptions on x, y and with the notations

above,

H0(F,X) = 0,

or qpx = 0. Moreover, sX ∩ Y ∼= Fp ⊂ X[s, p] in all cases, while if

H0(F,X) 6= 0, the bi-torsion X[s, p] ∼= F2p and q′x ∈ X[s, p] \ sX.

Proof. Let N be a sufficiently large integer, such that pNx = 0 and letR = Z/(pN · Z), so X is an R[s]-module. We note that in R′ := R/(N ) themaximal ideal is (s), thus a principal nilpotent ideal, as one can verify from

the definition of the norm N =∑p−1

i=0 νi. Indeed, in R we have the identity

p = sp−1 · v(s) mod N , v(s) ∈ R×, so the image of p in R′ is a power of s,hence the claim. As a consequence, in any finite cyclic R′-module M wehave M [p, s] ∼= Fp.

We consider the two modules N1 = Zpx and N2 = Rsx, such that X =N1 + N2, and only N2 is an R-module, and in fact an R′- module, since itis annihilated by N 1. Note that qN1 is also annihilated by the norm; thiscovers also the case when qx = 0, so qN1 = 0. Since X = N1+N2, it followsthat X[p] = N1[p] +N2[p].

We always have that N2[s] ∼= Fp, since N2 is an R′-module and (s) is themaximal ideal of R′. Let K := X[s] = Ker (s : X → X) and assume that

1This proof is partially inspired by some results in the Ph. D. Thesis of Tobias Bembom[1]

VANISHING OF µ 13

H0 := H0(F,X) 6= 0; then the bi-torsion X[p, s] = X[s][p] is not Fp-cyclic.It follows that X[p, s] 6∼= N1 ∩ X[p, s] ∼= Fp. Let r = qpe = ord(x), withe ≥ 0. Note that

y0 := (q/p)y = qxu(s) + (q/p)sp−1 ∈ X[p, s],

for all values of e. If e = 0, then y0 = (q/p)sp−1 ∈ N2[p, s]×. But then

(q/p)sx 6= 0 and thus N1[p, s] = 0 so H(F,X) = 0 in this case. Assumethat e > 0 and y0 ∈ N2. Since y0 = qx + (q/p)f(s)(sx) and the last term,

y1 := (q/p)f(s)(sx) ∈ N2, it follows that qx = y0− y1 ∈ N2, so H(F,X) = 0

in this case too. It remains that if H(F,X) 6= 0, then y0 ∈ N1 \ N2 ande > 0.

In this case N1 ∩N2 = ∅ and X = N1 ⊕N2. We let (q/p)(pex− cy) = 0,with (c, p) = 1. Then there is some w ∈ X[q/p] with pex = cy + w, and thedecomposition X = N1⊕N2 induces a decomposition of the q/p-torsions, sowe may write w = a(q/p)+bv(s)sNx, with N > 0 and a, b ∈ {0, 1, . . . , p−1},v(s) ∈ R′×. Taking norms in the identity (pe − a(q/p))x = cy + bv(s)sNxwe obtain

py · (c− pe−1 + a(q/p2)) = 0.

We have assumed q > p2, so for e > 1 the cofactor of py is a unit, whichimplies py = 0, in contradiction with ord(y) = q > p2. Therefore e = 1 and

y = px+ v(s)sp−1x, and (q/p)sp−1x = 0.(12)

Consequently, if H(F,X) 6= 0, then ord(x) = qp and there is an N <(p− 1)k such that N2[s] = Fp · (sNx). This completes the proof.

�

We can now assume that H0(F,X) = 0 and sinceX is finite, the Herbrand

quotient vanishes and H1(F,X) = 0. There is thus an exact sequence ofZp[s]-modules

0→ X[p]→ X → X → X/pX → 0,

in which X/pX is cyclic generated by x and the arrow X → X is themultiplication by p map ·p.

Since H0(F,X) = 0, it follows that Ker (s : X → X)[p] ∼= Fp and X[p] isFp[s]-cyclic, as is X/pX. Using this observation, we provide now the proofof the Proposition.

Proof. There is some distinguished polynomial φ ∈ R with φ(s)X = 0 anddeg(φ) = p-rk(X). Indeed, let d = p-rk(X) and x ∈ X/pX be the imageof x, so (sib)i=0,d−1 have independent images in X/pX, by definition of therank and span X, as a consequence of the Nakayama Lemma. Thereforesdx ∈ pX and there is a monic distinguished annihilator polynomial

φ(s) = sd − peh(s)of X, with e ≥ 0 and h a polynomial of deg(h) < deg(φ) ≤ p, which isnot p-divisible. Note that, by minimality of d, the case e = 0 only occurs

14 PREDA MIHAILESCU

if φ(s) = sd. We shall distinguishe several cases depending on the degreed = deg(φ) and on properties of h(s).

We consider the cases in which d < p first. Assume that scx = 0 for somec < p, so h = 0. Upon multiplication with sp−1−c we obtain sp−1X = 0.Thus N (x) = y = (sp−1 + pu(s))x = pu(s)x, and yu−1(s) = y = px, whichsettles this case.

We can assume now that φ(s) = sd − peh(s) and e > 0, with h(s) a nontrivial distinguished polynomial of degree deg(h) < d < p. As a consequence,if d < p− 1, then

y = N (x) = pu(s)x+ sp−1x = p(u(s) + sp−1−dpe−1h(s))x

= pv(s)x, v(s) = u(s) + sp−1−dpe−1h(s) ∈ R×.

Hence y = v−1(s)z = px, which confirms the claim in this case.If d = p− 1, then sp−1x = p · (pe−1h(s))x and thus

y = (pu(s) + sp−1)x = p(u(s) + pe−1h(s))x;

if the expression in the brackets is a unit, we may conclude like before.Otherwise, e = 1 and h(s) = −1+sh1(s), and thus y ∈ sX, so N (y) = py =0, in contradiction with the assumption2 that ord(y) > p.

The case d = p is more involved. We claim that ord(x) < p2ord(y).Assume that this is not the case. Since d = p, we have p-rk(X) = p and thussp−1x = y − pxu(s) 6∈ pX. In particular y 6∈ pX. Let q = pk = ord(y) andassume that ord(x) = pe+k = pe · ord(y), e > 1. We note that ord(sp−1x) =pe−1q, since sp−1u−1(s)x = −px + y has annihilator pe−1q. Consider thegenerators sjx of X; there is an integer j in the interval 0 ≤ j < p− 1, suchthat

ord(sjx) = ord(x) > ord(sj+1x) = ord(sp−1x) = ord(x)/p.

Recall from Lemma 3.6, that we are in the case when X[p] is a cyclic Fp[s]

module of dimension p as an Fp-vector space, and H0(F,X) = 0. Let

F0 := {qpe−1six : i = 0, 1, . . . , j} ⊂ X[p],

F1 := {qpe−2sj+ix : i = 1, 2, . . . , p− j − 1} ⊂ X[p],

and F = F0 ∪ F1. Then Fi ⊂ X[p] are Fp-bases of some cyclic Fp[s] sub-modules F0, F1 ⊂ X[p] with dimFp(F0) ≤ j+1 and dimFp(F0) ≤ p− (j+1).We claim that X[p] = F0⊕F1 = 0. For each z ∈ X[p] there is some maximalz′ ∈ X – thus z′ having non-trivial image 0 6= z′ ∈ X/pX – and such thatz = q′z′ for some q′ ∈ pN. Since the generators of X/pX are mapped thisway to F = F0+F1, which is an Fp-vector space, it follows by linearity thatF = X[p]. Comparing dimensions, we find that F0 ∩ F1 = 0 so there is adirect sum F = F0 ⊕ F1 = X[p], as claimed.

2At this point the assumption that ord(y) > p plays a crucial role, and if it were notto hold, modules such that ord(x) becomes arbitrarily large are conceivable

VANISHING OF µ 15

Note that

0 6= (q/p)y = qx+ (q/p)sp−1u−1(s)x ∈ X[p][s];

upon multiplication with pe−1 ≥ p we obtain

0 = qpe−2y = qpe−1u(s)x+ qpe−2sp−1x

= qpe−1(u0(s) + u1(s))x+ qpe−2sp−1x,(13)

where u0 ∈ F0 and u1 ∈ F1 are the projections of the unit u(s), thus

u0(s) ≡∑j

i=0 si(pi

)

pmod p, u1(s) ≡

∑p−2i=j+1 s

i(pi

)

pmod p.

By definition of j, it follows that

f0 := qpe−1u0(s)x ∈ F0, f1 := sp−1qpe−2x ∈ F1,

and qpe−1u1(s)x = 0. The identity in (13) becomes f0 + f1 = 0, and sinceF0 ∩ F1 = 0 and fi ∈ Fi, i = 0, 1, it follows that f0 = f1 = 0. However,f1 = sp−1qpe−2x ∈ F1 is a basis element which generates F1[s], so it cannotvanish. The contradiction obtained implies that we must have in this casealso ord(x) ≤ pq. Consequently, ord(x) ≤ p · ord(y) in all cases, whichcompletes the proof of the Proposition. �

3.4. The Norm Principle, ray class fields and proof of Lemma 2.2.

Let q be a rational prime with q ≡ 1 mod pm. Let F ⊂ Qq[ζq] be the subfieldof the (ramified) q-th cyclotomic extension, which has degree p over Qq.Thus F is the completion at the unique ramified prime above q of the field F

defined in the text. The field F is a tamely ramified extension of Qq, so classfield theory implies that Gal(F/Qq) is isomorphic to a quotient of order p of(Z/q · Z)∗, so letting S = ((Z/q · Z)∗)p, we have Gal(F/Qq) ∼= (Z/q · Z)∗/S.Let thus r ∈ Z be such that r(q−1)/p 6≡ 1 mod q: if g ∈ F×

q generates themultiplicative group of the finite field with q elements and m = vp(q − 1),

then one can set r = g(q−1)/pm rem q. Let in addition Kn/Qq be the pn-th

cyclotomic extension of Qq: under the given premises, we have for n > m the

extension degree [Kn : Qq] = pn−m and Kn = Qq

[r1/p

n−m], while for n ≤ m

the extension is trivial. Letting rn = r1/pn−m ∈ Kn, and Ln = Kn · F, we

deduce by class field theory that rn generates K×n /N (L×

n ), under the naturalprojection. Indeed, the extension Ln/Kn is a ramified p extension, so thegalois group must be a quotient of the roots of unityW (Zq), hence the claim.We shall use these elementary observation in order to derive the structure

of H1(F,A−(Ln)) in our usual setting and prove some necessary conditions

for elements vn ∈ A−(Ln) which verify 0 6= β(vn) ∈ H1(F,A−(Ln)), under

the natural projection β : A−(Ln)→ H1(F,A−(Ln)).Let K,L,F, etc. be the fields defined in the main part of the paper and let

us denote by I(M) the ideals of some arbitrary number field, and P (M) ⊂I(M) the principal ideals. The maximal p-abelian unramified extension isH(M) and the p-part of the ray class field to the ray Mq = q · O(M) will

16 PREDA MIHAILESCU

be denoted by Tq(M). If M is a CM field, then complex conjugation acts,inducing I−(M), P−(M) and H−,T−

q , in the natural way. In our context, we

let in addition P−N (L) := N (P−(L)) ⊂ P−(K). We let ∆n = Gal(Kn/k) for

arbitrary n. The following is an elementary result in the proof of the HasseNorm Principle:

Lemma 3.7. Let K,L and F be like above. Then

H(1)(F,A−(Ln)) ∼= P−(Kn)/P−N (Ln) ∼= Fp[∆m], for all n > 0,(14)

the isomorphism being one of cyclic Fp[∆m]-modules.

Proof. Note that both modules in (14) are annihilated by p. In the case ofP−(K)/P−

N (L), this is a direct consequence of (P−(K))p = NL/K(P−(K)) ⊂

P−N (L). Let β : A−(L)→ H(1)(F,A−(L)) and πN : P−(K)→ P−(K)/P−

N (L)denote the natural projections and let a ∈ Ker (N : A−(L) → A−(L)).Then pu(s)a = −sp−1a and thus pa = −sp−1u−1(s)a and a fortiori β(pa) = 0

for all a, so pH(1)(F,A−(L)) = 0, thus confirming that H(1)(F,A−(Ln)) isan Fp-module too.

Let now A ∈ a be some ideal and (α) = N (A). The principal ideala := (α/α) ∈ P−(K) has image πN (a) ∈ P−(K)/P−

N (L) which dependson a but not on the choice of A ∈ a. This is easily seen by choosing adifferent ideal B = (x)A ∈ a: then N (B1−) = a · N (x/x) ∈ a · P−

N (L),and πN (N (B1−)) = πN (N (A1−)) = πN (a) depends only on a. Supposenow that a ∈ P−

N (L), so πN (a) = 1. Then there is some y ∈ L× such that

N (A1−) = (N (y)1−) and thus N (A/(y))1− = (1). Since H(1) vanishes forideals, it follows that there is a further ideal X ⊂ L such that

A1− = ((y)Xs)1− ,

and thus a2 ∈ (A−(L))s. But then β(a) = 0. We have shown that there is a

map λ : H(1)(F,A−(L)) → P−(K)/P−N (L) defined by the sequence of asso-

ciations β(a) 7→ a 7→ πN (a), which is a well defined injective homomorphismof Fp-modules.

In order to show that λ is an isomorphism, let x := (x/x) ∈ P−(K)\P−N (L)

be a principal ideal that is not a norm from L. Let the Artin symbol of x be

σ =(L/Kx

)∈ Gal(L/K); by definition, L is also CM , so complex conjugation

commutes with σ and we have(L/K

x

)=

(L/K

x

)= σ = σ.

Consequently,(

L/K(x/x)

)= 1 – we may thus choose, by Tchebotarew, a prin-

cipal prime (ρ) ⊂ K with ρ ∼= x/x mod q, and which is split in L/K. LetR ⊂ L be a prime above (ρ) and r := [R1−] ∈ (A−)(L). We claim thatβ(r) 6= 0; assume not, so R = (y)Ys for some y ∈ L and Y ⊂ L and thusN (R1−) = (N (y/y)) ∼= x mod P−

N (L). Since (N (y/y)) ∈ P−N (L) by defini-

tion, it follows that x = (x/x) ∈ P−N (L), which contradicts the choice of x.

VANISHING OF µ 17

It follows that λ is an isomorphism of Fp[∆m]-modules. The proof will becompleted if we show that P−(K)/P−

N (L) ∼= Fp[∆m]. Since ∆m acts transi-tively on the pairs of complex conjugate primes above q in Kn, one verifiesthat (K×

n )−/N ((L×

n )−) ∼= Fp[∆m]. The field k contains no p-th roots of

unity, and thus NKn/k(P−(Kn)/P

−N (Ln)) 6= 0. The isomorphism

P−(K)/P−N (L) ∼= Fp[∆m]

follows, and this completes the proof. �

Consider the field T′n := T−

q (Kn) defined as the minus p-part of the rayclass field to the modulus Qn := qO(Kn) – i.e., if T is the full ray class fieldto Qn and Tp is the p-part of this field, thus fixed by all q′-Sylow subgroupsof Gal(T/Kn), for all primes q′ ≤ q, then T′

n is the subfield of Tp fixed by(Gal(Tp/Kn))

+. As mentioned above, the completions at individual primesq′ above q are cyclic groups

T′n,q′/H(Kn)q′ ∼= (Fq[ζpn ]

×)p ∼= Cpn .

Consequently,

Gal(T′n/H(Kn)) ∼=

∏

g∈∆m

Cpn .(15)

Since q is ramified in Ln/Kn, the residual fields of the ray class subfieldTn := T−

q (Ln) are the same as the ones of T′n and thus Tn = H(Ln) ·T′

n. Letalso E′

n ⊂ T′n be the maximal subextension with pGal(E′

n/H′n) = 0 – thus

the p-elementary extension of H(Kn) contained in T′n – and En = E′

n ·H(Ln).Since the local extensions T′

n,q′/H(Kn)q′ are cyclotomic, the ramification of

E′n/H(Kn) is absorbed by Ln/Kn and therefore En ⊂ H(Ln).

Consider a class x ∈ A−(Ln) such that 0 6= β(x) ∈ H1(F,A−(Ln)) andlet A ∈ x be a prime and (α) = N (A) ⊂ Kn be the principal ideal belowit. By Lemma 3.7, it follows that πN ((α/α)) 6= 0 and thus the Artin sym-

bol y′ =(T′

n/Kn

(α/α)

)∈ Gal(T′

n/Kn) generates a cycle of maximal length in

Gal(T′n/H(Kn)), so it acts non trivially in E′

n/H(Kn). If y ∈ Gal(Tn/Ln)is any lift of ϕ(x), i.e. y|H(Ln) = ϕ(x), then N (y′) acts non trivially in

E′n/H(Kn). The converse holds too. Suppose that x ∈ A−(Ln) and let

y ∈ Gal(Tn/Ln) be some lift of ϕ(x). If N (y) fixes H(Kn) and acts nontrivially in E′

n/H(Kn), then β(x) 6= 0. Indeed, by choosing A ∈ x a prime

with(Tn/Kn

A

)= y, we see that I := N (A) must be a principal ideal, since

(Tn)/Kn

I

)fixes H(Kn) and moreover, the Artin symbol acts non trivially on

E′n/H(Kn), so πN (I) 6= 0. The claim follows from Lemma 3.7.Let T = ∪nTn. In the projective limit, we conclude that x = (xn)n∈N has

β(x) 6= 0 iff for any lift y ∈ Gal(T/L∞) of ϕ(x) ∈ Gal(H(L∞)) the normN (y) fixes H(K∞) and acts non trivially in E′/H(K∞), with E = ∪nE′

n.Moreover, T/H(L∞) is the product of |∆m| independent Zp-extensions and

18 PREDA MIHAILESCU

Q

K L

Kn Ln

H′n

•

E′n

Hn

T′n

Tn

K∞ L∞

p

p

N (y) is of λ-type. We have thus from 14 a further isomorphism,

H1(F,A−(Ln)) ∼= Gal(E′n/H(Kn)) ∼= Gal(E′

∞/H(K∞).(16)

We have thus proved:

Fact 3.8. For every n > m, a class x ∈ A−(Ln) has non trivial im-

age β(x) ∈ H1(F,A−(Ln)) iff for any lift y ∈ Gal(Tn/Ln) of ϕ(x) ∈Gal(H(Ln)/Ln), the norm y′ := N (y) ∈ Gal(T′

n/Kn) fixes Hn and acts non

trivially in E′n; equivalently, y

′ generates a maximal cycle in Gal(T′n/H(Kn)).

In the projective limit, x = (xn)n∈N has β(x) 6= 0 iff for any lift y ∈Gal(T/L∞) of ϕ(x) ∈ Gal(H(L∞)) the norm N (y) fixes H(K∞) and acts

non trivially in E′/H(K∞), with E = ∪nE′n. Moreover, there is an exact

sequence

1→ Gal(H(L∞/L∞)→ Gal(T/L∞)→ (Zp)|∆m| → 1,(17)

and thus Gal(T/K∞) is Noetherian Λ-module.

We now prove the Lemma 2.2:

Proof. Assume that H1(F,

(M(pA−(L))

))6= 0. Consider the modulesMn =

Gal(Tn/Ln) and M = lim←−nMn = Gal(T/L∞). It follows from Fact 3.8 that

M/Gal(H(L∞)) ∼= Z|∆m|p , so M is a Noetherian Λ-module and the exact

VANISHING OF µ 19

sequence (17) shows that M is a rigid module and the further premises ofProposition 1.3 hold too, as a consequnece of the choice of K. Let v′ ∈M besuch that the restriction v = v′

∣∣H(L∞)

∈ M(pA−(L)) and it has non trivial

image in H1(F,

(M(pA−(L))

)), via the inverse Artin map. In particular v 6∈

L(pA−(L)) and also v′ 6∈ L(M). Assume that ord(v) ≤ p exp(M(pA−(K))).The Proposition 1.3 together with Remark 1.4 imply that T 2v′ = v′µ + v′λis decomposed. But then v′µ ∈ Gal(H(L∞)) so it follows from the above

Fact that β(T 2v) = 0. Thus, if v ∈ M(pA−(L)) has order ord(v) ≤ p ·exp(M(pA−(K)), then T 2v ∈ sM(pA−(L)), which completes the proof ofLemma 2.2. �

3.5. Proof of the Proposition 1.3. The proof of the Proposition requiresa longer analysis of the growth of modules Λxn for indecomposed elements.This will be divided in a sequence of definitions and Lemmata which even-tually lead to the proof.

The arguments of this section will take repeatedly advantage of the fol-lowing elementary Lemma3:

Lemma 3.9. Let A and B be finitely generated abelian p−groups denoted

additively, and let N : B → A, ι : A→ B be two Zp - linear maps such that:

1. N is surjective.

2. The p−ranks p-rk(A) = p-rk(pA) = p-rk(B) = r.3. N(ι(a)) = pa,∀a ∈ A.

Then

A. The inclusion ι(A) ⊂ pB holds unconditionally.

B. We have ι(A) = pB and B[p] = Ker (N) ⊂ ι(A). Moreover,

ord(x) = p · ord(ι(N(x))) for all x ∈ B.

C. If there is a group homomorphism T : B → B with ι(A) ⊆ Ker (T )and ν := ι ◦N = p +

(p2

)T + O(T 2), then ν = ·p, i.e. ι(N(x)) = px

for all x ∈ B.

Proof. Since A and B have the same p-rank and N is surjective, we knowthat the map N : B/pB → A/pA is an isomorphism4. Therefore, the mapinduced by Nι on the roof is trivial. Hence ι : A/pA → B/pB is also zeroand thus ι(A) ⊂ pB, which confirms the claim A.

The premise p-rk(pA) = p-rk(A) implies that p-rk(A) = p-rk(ι(A)), asfollows from

p-rk(A) ≥ p-rk(ι(A) = p-rk(ι(A)[p]) ≥ p-rk(pA) = p-rk(A).

We now consider the map ι′ : A/pA→ pB/p2B together with N . From thehypotheses we know that Nι′ is the multiplication by p isomorphism: ·p :

3I owe to Cornelius Greither several elegant ideas which helped simplify my originalproof.

4For finite abelian p-groups X we denote R(X) = X/pX by roof of X and S(X) = X[p]is its socle

20 PREDA MIHAILESCU

A/pA → pA/p2A, as consequence of p-rk(A) = p-rk(ι(A)) = p-rk(pA). Itfollows that ι′ is an isomorphism of Fp-vector spaces and hence ι : A→ pB issurjective, so ι(A) = pB. Consequently |B|/|ι(A)| = pr. Let x ∈ Ker (N);since N : B/pB → A/pA is surjective, the norm does not vanish for x 6∈ pB.Consequently, for x ∈ Ker (N) ⊂ pB we have Nx = px = 0, so Ker (N) =B[p] ⊂ ι(A), so Ker (N) = ι(A)[p], as claimed.

We now prove that ord(x) = p · ord(ι(N(x)) for all x ∈ B. Consider thefollowing maps π : B → ι(A), x 7→ px and π′ = ι ◦ N . Since pB = ι(A),both maps are surjective and there is an isomorphism φ : pB → pB suchthat π = φ ◦ π′. Therefore

ord(x)/p = ord(px) = ord(φ(px)) = ord(ι(N(x))),

and thus ord(x) = p · ord(ι(N(x))), as claimed.For point C. we let x ∈ B, so px ∈ pB = ι(A) and thus pTx = T (px) = 0.

Consequently Tx ∈ B[p] ⊂ ι(A) and thus T 2x = 0. From the definition of

ν = ι ◦ N = p + Tpp−12 + O(T 2) we conclude that νx = px + p−1

2 Tpx +

O(T 2)x = px, which confirms the claim C. and completes the proof. �

In this section M∞/M is an arbitrary Zp-extension in which all the primesthat ramify, ramify completely and X ′ ⊆ Y := Gal(T/M∞) is a NoetherianΛ-submodule, the limit of the ray class groups Gal(Tn/Mn) to some raymodule associated to the base field with trivial finite part. Thus ωnx = 0implies Tx = 0 by Fact 3.3. The ray class groups Yn = ϕ−1(Gal(H(Tn)/M∞)may also coincide with class groups A(Mn).

We shall apply the Theorem 1.3 to the concrete cases in which M = L

and Y is either Gal(H−(L)/L∞) or Gal(T/M∞), where T is the injectivelimit of subfields of ray class groups, defined above.

We denote by L,M,D the λ-, the µ- and the decomposed parts, re-spectively, of X ′: thus, in the notation of the introduction, we have L =L(X ′),M =M(X ′),D = D(X ′). One can for instance think of M as K in§2 and X ′ = ϕ(pA−(K)) = X− or X ′ = Gal(T/H(K∞)).

For x ∈M , the order is naturally defined by ord(x) = min{pk : pkx = 0}.Since X ′ has no finite submodules, it follows that ord(x) = ord(T jx) for allj > 0.

We introduce some distances dn : X ′ ×X ′ → N as follows: let x, z ∈ X ′;then

dn(x, z) := p-rk(Λ(xn − zn)); dn(x) = p-rk(Λxn).

We obviously have dn(x, z) ≤ dn(x, y)+dn(x, z) and dn(x) ≥ 0 with dn(x) =0 for the trivial module. Also, if f ∈ Zp[T ] is some distinguished polynomialof degree φ = deg(f), then dn(x) − φ ≤ dn(fx) ≤ dn(x) for all x ∈ X. Weshall write d(x, y) = limn dn(x, y). Also, for explicit elements u, v ∈ X ′

k, wemay write d(u, v) = p-rk(Λ(u− v)). This can be useful for instance when noexplicit lifts of u, v to X ′ are known. The simplest fact about the distanceis:

VANISHING OF µ 21

Fact 3.10. Let x, z ∈ X ′ be such that dn(x, z) ≤ N for some fixed bound

N and all n > 0. Then x − z ∈ L and N ≤ ℓ := p-rk(L). For every fixed

d ≥ p-rk(L) there is an integer n0(d) such that for any x ∈ X ′ \ L and

n > n0, if dn(x) ≤ d then x ∈ νn,n0X ′.

Proof. The element y = x − z generates at finite levels modules Λyn ofbounded rank, so it is neither of µ-type nor indecomposed. Thus y ∈ L andconsequently dn(y) ≤ p-rk(Ln) ≤ ℓ for all n, which confirms the first claim.

For the second claim, note that if x 6∈ L, then dn(x)→∞, so the bound-edness of dn(x) becomes a strong constraint for large n. Next we recall thatF (T )x ∈M and since dn(F (T )x) ≤ dn(x), we may assume that x ∈M . Nowdn(x) ≤ d implies the existence of some distinguished polynomial h ∈ Zp[T ]with deg(h) = d and such that h(T )xn = 0. The exponent of M is boundedby pµ, so there is a finite set H ⊂ Zp[T ] from which h can take its values.Let now n0 be chosen such that νn0,1 ∈ (h(T ), pµ) for all h ∈ H. Such a

choice is always possible, since νn,1 = pµ · V (T ) + T pn−µ−1 ·W (T ) for someV (T ),W (T ) ∈ Λ. We may thus choose n sufficiently large, such that the

Euclidean division T pn−µ= q(T )h(T ) + r(T ) yields remainders r(T ) which

are divisible by pµ for all h ∈ H. Let n0 be the smallest such integer. Withthis choice, for any h ∈ H, it follows that h(T )xn = 0 implies νn0,1xn = 0and thus νn0,1x = νn,1w for some w ∈ X ′. Then νn0,1(x − νn,n0

w) = 0 andthus x = νn,n0

w, by Fact 3.3 and the assumption on X ′.�

We pass now to the proof of Proposition 1.3.

Proof. Let x ∈ X ′ and suppose that l is the smallest integer such thatplx ∈ L and let fx(T ) be the minimal annihilator polynomial of plx, soy := fx(T )x ∈M , since ply = 0. We claim that

pjxn+j − ιn,n+j(xn) ∈ fx(T )Λxn+j ⊂ Λyn+j, ∀j > 0,

and in particular ιn,n+l(xn) = plxn+l − hn+l(T )(fx(T )xn+l) is decomposed.We let n1 > n0 be such that for all x = (xn)n∈N in X ′ \ pX ′ we haveord(pl+µxn) > p; here n0 is the constant established in Fact 3.10 with respectto the bound rank d = p-rk(L) + 1. For n > n1 and x ∈ X ′ \ (D + pX ′), wehave

plxn+l − ιn,n+l(xn) = fx(T )hn(T )xn+l ∈Mn+l.(18)

Indeed, consider the modulesB = Λxn+1/(fx(T )·Λxn+1) and A = Λxn/(fx(T )·Λxn). Since ιn,n+1xn 6∈ fx(T )Λxn+1 for n > n1 – as follows from the condi-tion imposed on the orders – the induced map ι : A→ B is rank preserving.We can thus apply the Lemma 3.9, an l-fold iteration of which implies theclaim (18). We now apply the hypothesis that px = c + u ∈ D for somec ∈ L, u ∈ M [pl−1]; note that the condition (4) allows us to conclude fromLemma 3.9 and c ∈ L that ιn,n+k(cn) = pkcn+k for all c = (cn)n∈N ∈ L In

22 PREDA MIHAILESCU

particular,

pωncn+1 = ωnιn,n+1(cn) = 0, so ωn+1cn+1 ∈ Ln+1[p] = ι1,n+1(L1[p]).

as a consequence of the same Lemma. We deduce under the above hypothesison n, that

plxn+l = pl−1cn+l = ιn+1,n+lcn+1 = ιn,n+l(xn) + hyn+l.

By applying ωn to this identity and using ωncn+1 ∈ ι1,n+1(L1[p]), so Tωncn+1 =0, we find Th(T )ωnyn+l = 0. The Iwasawa’s Theorem VI [5] implies thatthere is some z(n) ∈ X ′ such that Thωny = νn+l,1z(n). Since ply = 0, we

have in addition plνn+l,1z(n) = 0, so plz(n) = 0 and z(n) ∈M , by Fact 3.4.We stress here the dependency of z ∈ X ′ on the choice of n > n1 by writingz(n); this is however a norm coherent sequence and zm(n) will denote itsprojection in X ′

m for all m > 0. We obtain νn,1(T2h(T )y − νn+l,nz(n)) = 0.

This implies T 2h(T )y = νn+l,nz(n), by the same Fact 3.4. Reinserting thisrelation in the initial identity, we find

ιn,n+l(T2xn − zn(n)) = ιn+1,n+l(T

2cn+1).(19)

We prove that (19) implies that T 2x must be decomposed. For this we

invoke the Fact 3.10 with respect to the sequence w(n) = T 2x − z(n). Wehave dn(ιn+1,n+l(T

2cn+1)) ≤ p-rk(L) for all n; since z(n) ∈ M , and T 2x is

assumed indecomposed, then w(n) 6∈ D either, and the Fact 3.10 together

with the choice of n1 imply that w(n)n ∈ νn,n0

X ′ for all n > n1. But then

w(n)n = ιn,n+l(T

2xn − zn(n)) = νn,n0an ∈ ιn0,n(X

′n0).

It follows in particular that

ord(T 2xn − zn(n)) ≤ plord(ιn,n+l(T2xn − zn(n))) ≤ pl exp(X ′

n0).

This holds for arbitrary large n, and since zn(n) ∈Mn, we have ord(T2xn−

zn(n)) = ord(T 2xn) for n > n1. Therefore, the assumption that Tx 6∈ Dimplies that ord(T 2xn) ≤ pl exp(X ′

n0) for all n > n1: it is thus uniformly

bounded for all n, which would imply that x ∈M , in contradiction with theassumption x 6∈ D. We have thus proved the claim for all x ∈ X ′\(D+pX ′).The general case follows by applying Nakayama to the module X ′. �

Acknowledgments: The main proof idea was first investigated in thewish to present the result in the memorial volume [2] for Alf van der Poorten.However not more than the trivial case was correctly proved; Soren Kleineundertook within part of his PhD Thesis the task of brushing up the proof.He succeeded to do so, with the exception of the details related to decom-position and Lemma 2.2, which we treated here. I owe to Soren Kleinefor numerous useful discussions sustained during the development of thispaper. I am particularly grateful to Andreas Nickel for critical discussionsof an earlier version of this work. In Gottingen, the graduate students inwinter term of 2015-2016, Pavel Coupek, Vlad Crisan and Katharina Mullerpresented in a seminar this work, I owe them for their great contribution to

VANISHING OF µ 23

the verification of the final redaction. The paper is dedicated to the memoryof Alf van der Poorten.

References

[1] T. Bembom. The Capitulation Problem in Class Field Theory. PhD thesis, GeorgAugust Universitat Gottingen, April 2012.

[2] J. Borwein, I. Shparlinski, and W. Zudilin, editors. Number Theory and Related Fields:In Memory of Alf van der Poorten. Springer, 2013.

[3] B. Ferrero and L. Washington. The Iwasawa invariant µp vanishes for abelian numberfields. Annals of Mathematics (2), pages 377–395, 1979.

[4] K. Iwasawa. Number Theory, Algebraic Geometry and Commutative Algebra, chapterOn the µ-invariants of cyclotomic fields, pages 1–11. Kinokunia, Tokio, 1973. In honorof Y. Akizuki.

[5] K. Iwasawa. On Zℓ - extensions of number fields. Ann. Math. Second Series, 98:247 –326, 1973.

[6] J. Sands. On small Iwasawa invariants and imaginary quadratic fields. Proccedings ofthe American Mathematical Society, 112(3):671–684, 1991.

(P. Mihailescu) Mathematisches Institut der Universitat Gottingen

E-mail address, P. Mihailescu: preda@uni-math.gwdg.de